Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $n = \dfrac{-5x + 35}{x^2 - 15x + 50} \times \dfrac{-3x^2 + 18x - 15}{x - 1} $
First factor out any common factors. $n = \dfrac{-5(x - 7)}{x^2 - 15x + 50} \times \dfrac{-3(x^2 - 6x + 5)}{x - 1} $ Then factor the quadratic expressions. $n = \dfrac {-5(x - 7)} {(x - 5)(x - 10)} \times \dfrac {-3(x - 5)(x - 1)} {x - 1} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {-5(x - 7) \times -3(x - 5)(x - 1) } { (x - 5)(x - 10) \times (x - 1)} $ $n = \dfrac {15(x - 5)(x - 1)(x - 7)} {(x - 5)(x - 10)(x - 1)} $ Notice that $(x - 5)$ and $(x - 1)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {15\cancel{(x - 5)}(x - 1)(x - 7)} {\cancel{(x - 5)}(x - 10)(x - 1)} $ We are dividing by $x - 5$ , so $x - 5 \neq 0$ Therefore, $x \neq 5$ $n = \dfrac {15\cancel{(x - 5)}\cancel{(x - 1)}(x - 7)} {\cancel{(x - 5)}(x - 10)\cancel{(x - 1)}} $ We are dividing by $x - 1$ , so $x - 1 \neq 0$ Therefore, $x \neq 1$ $n = \dfrac {15(x - 7)} {x - 10} $ $ n = \dfrac{15(x - 7)}{x - 10}; x \neq 5; x \neq 1 $